BPCF 2019 - XIV Београдски фестивал проблемског шаха 2019.
Belgrade Internet Tournaments 2019
Group C – fairy hs#3, Protean chess
Judge:
Petko Petkov, International Judge
Director:
Marko Ložajić
This year's BIT tournament (section C) is dedicated to the
30th anniversary of the death of legendary grandmaster Nenad Petrović. I am very pleased to note
that the level of the tournament is high; a number of very interesting problems have been presented,
which will certainly stimulate future work on this theme!
Nenad Petrović
(7th Sep 1907 – 9th Nov 1989)
Thanks to all the participants, and congratulations to the winners!
The theme of this tournament was:The fairy conditions Protean Chess or Frankfurt Chess, presented in HS#3
with exactly 2 solutions.
Above all, I must point out that these two conditions are almost identical, and the general requirement for them is:
"When a piece captures (King included), it takes the nature of the captured unit (without changing colour).
A King capturing becomes a royal unit". (Nowadays this definition exists in WinChloe and the condition is
denoted by the French name Francfort).
In fact, this is the exact definition of Frankfurter Chess, first introduced by the inventors Albert Heinrich
Kniest & John Niemann in The Fairy Chess Review, Vol. 7, Nr. 1, VIII / 1948. The name of the condition, however,
was coined by Paul Schlensker.
The predecessor of Protean Chess is the Protean piece, also defined by A. H. Kniest around 1940 in
Unsere Schachwelt, but historically it is not known when exactly the Protean Chess condition was established.
The difference between Frankfurter Chess and Protean Chess is the following: If an enemy pawn is captured in
Protean Chess the new pawn on the same square (with a changed color) moves in the direction of the captured pawn.
In Frankfurter Chess, of course, this strange rule is missing (nowadays Protean Chess exists in Popeye).
When there are no pawn captures or these captures have no thematic effect and do not affect the correctness of
the problem, we can assume that the two conditions are the same. In this case, the author may indicate below the
diagram what condition he wishes.
For the sake of clarity, I decided to place the following additional symbols after the solutions of the problems,
using the first letters of the program names Popeye (P) and WinChloe (W): (P + W) means that it is possible to use
either of the two conditions; (P) means that only Protean Chess is used; (W) means that only Frankfurter Chess is used.
From the tournament director I received 24 originals by 18 authors from 11 countries. I eliminated from the tournament
on various grounds the following problems: No: 17, 21, 24 these problems do not meet the regulation as they have twins
or set-play. No 4 - Not optimal construction. It is easily possible to get a position with 2 pieces less if you move
the starting position with one vertical to the right. No 11 - The wBc1 is superfluous in the solution after 1.g8 = Q.
I suggest the following ranking
Thematic condition
HS#3, 2 solutions, Protean Chess
No other fairy conditions or fairy pieces are allowed, except royal pieces (only one for each side!): Royal Queen (rQ), Royal Rook (rR), Royal Bishop (rB), Royal Knight (rS) and Royal Pawn (rP). These royal pieces may be present in the initial position (example 2) or may appear during the play. Royal Pawn promotions are allowed. "Supernumerary" pieces in initial position (e.g. 2 white Queens, 3 white Bishops, 3bR, 4 bS, etc...) are permitted if this form is thematically necessary for the problem.
Protean Chess: When a piece (King included) captures, it takes the nature of the captured unit (without changing color). A King capturing becomes a royal unit.
Royal Piece: Piece that executes a function of the King on the board: it becomes object of check and mate.
Example 1
Petko PETKOV
Original
white Pe4f5 Kd8 Be8 Sg8 Rf1
black Pc6g4h4 Sd7a6 Bf8d5 Kg5 Rd2
hs#3
2.1.1.1.1.1
(6+9)
Protean chess
1.Rf1-c1 Sa6-c5 2.Rc1*c5=S Sd7-b6 {
}
3.Sc5-e6 + Bd5*e6=S #
1.f5-f6 Bf8-g7 2.f6*g7=B Bd5-e6 {
}
3.Bg7-f6 + Sd7*f6=B #
Thematic sacrifices of Sa6 & Bf8, reciprocal play of Sd7 & Bd5 exchanging their nature: B becomes S and S becomes B
Example 2
Petko PETKOV
Original
white Kf2 Sc4
black Qa4 Sf3
black Royal Pf5
hs#3
2.1.1.1.1.1
(2+3)
Protean chess
Royal Pf5
1.Sc4-d2 Sf3-e1 2.Kf2*e1=rS Qa4-e4 + {
} 3.Sd2*e4=Q + rPf5*e4=rQ #
1.Sc4-e5 Sf3-g1 2.Kf2*g1=rS Qa4-g4 + {
} 3.Se5*g4=Q + rPf5*g4=rQ #
Multiple thematic sacrifices lead to surprising Ideal Echo-mates. Tanagra form.
Winners
Ofer Comay
Ivo Tominić
Kostas Prentos
Mario Parrinello
Participants
Themis Argirakopoulos – GRE (22)
S.K. Balasubramanian – IND (2*,15)
Ofer Comay – ISR (1)
Vlaicu Crisan - ROU (2*)
Borislav Gadjanski – SRB (16)
Eric Huber – ROU (17,24)
Igor Kochulov – RUS (3,9)
Juraj Lörinc – SVK (4,10)
Manikumar S – IND (13,14)
Luis Miguel Martín – ESP (18)
Miodrag Mladenović – SRB (19)
Karol Mlynka – SVK (5,11)
Mario Parrinello – ITA (6)
Kostas Prentos – USA (7)
Sergej Smotrov– RUS (20)
Viktor Syzozenko – UKR (21)
Ivo Tominić – CRO (23)
Aleksandr Tyunin – RUS (8,12)
18 names
from 11 countries
BPCF2019C - Award
1st Prize
Ofer COMAY, ISR
white Pb4d2f2g2 Se8 Bb3 Kc1 Re2
black Bc5 Kf8 Qd4 Pb6g3g5 Sc2d1 Rf4
hs#3
2 sol.
(8+9)
Protean Chess
1.f2*g3=PP Qd4-f2 2.Kc1*d1=rS Rf4-d4 3.rSd1*f2=rQ + Rd4-f4 #
1.b4-b5 Qd4-b4 2.Kc1*c2=rS Bc5-d4 3.rSc2*b4=rQ + Bd4-c5 #
A remarkable problem presenting a very interesting and difficult thematic complex. With their W1 moves, the white pawns vacate squares for the black Queen, which sacrifices itself in order to attract the white king after the W2 moves to the f2/b4 squares in Queen phases! Further, the formation of reciprocal black batteries R/B and B/R takes place in a very interesting way - with the opening of lines, then closing of the same lines in combination with Switchbacks. Of course, the active presence of the Umnov theme is also a very fine moment.
FEN: 4Nk2/8/1p6/2b3p1/1P1q1r2/1B4p1/2nPRPP1/2Kn4
(C+ by Popeye + WinChloe)
2nd Prize
Ivo TOMINIĆ, CRO
white Rf7 Sf8 Bb8d5 Kg3 Ph6
black Rh2 Kf5 Bf6f3
hs#3
2 sol.
(6+4)
Protean Chess
1.Kg3*h2=rR Bf3-g4 2.Bd5-f3 Bg4-h5 3.rRh2*h5=rB Kf5-g5 #
1.Kg3*f3=rB Rh2-h4 2.Bb8-h2 Rh4-h5 3.rBf3*h5=rR + Kf5-g4 #
Although with only 10 pieces, the author presents a very non-standard content. Two duets of pieces: Rh2/Bf3 and Bd5/Bb8 reciprocally change their functions. Particularly surprising and beautiful are the tempo-moves (B1) combined with the Umnov theme, white distant blocks on f3/h2, and diagonal-orthogonal Echo. This synthesis of ideas seems quite difficult to implement with such limited material and therefore in my opinion deserves a high ranking! Unfortunately not everything is "cloudless" here, for example the technical black Bf6.
FEN: 1B3N2/5R2/5b1P/3B1k2/8/5bK1/7r/8
(C+ by Popeye + WinChloe)
3rd Prize
Kostas PRENTOS, USA
white Qa5 Pa7 Ka6 Rd6 Bc4
black Ka8 Bb1 Sf4
hs#3
2 sol.
(5+3)
Protean Chess
1.Bc4-e2 Sf4*e2=B + 2.Rd6-d3 Bb1*d3=R 3.Qa5-d8 + Rd3*d8=Q #
1.Rd6-g6 Bb1*g6=R + 2.Bc4-e6 Sf4*e6=B 3.Qa5-d5 + Be6*d5=Q #
A very beautiful and economical albeit not very difficult to solve problem in which we see a paradoxical phenomenon: all white and black pieces (except the kings and Pa7) disappear from the board in the process of play, or change their nature and color! The reciprocal creation of black batteries is interesting and at the same time very instructive in terms of the future development of such ideas!
FEN: k7/P7/K2R4/Q7/2B2n2/8/8/1b6
(C+ by Popeye + WinChloe)
Spec. Prize
Mario PARRINELLO, ITA
white Pe6f6 Kc7 Sb5g3
black Pd6c6 Kd2 Bc3 Re2
hs#3
2 sol.
(5+5)
Protean Chess
1.Sg3*e2=R + Kd2*e2=rR 2.Kc7*c6=rPP rRe2*e6=rPP 3.Sb5-d4 + Bc3*d4=S #
1.Sb5*c3=B + Kd2*c3=rB 2.Kc7*d6=rPP rBc3*f6=rPP 3.Sg3-e4 + Re2*e4=S #
Another excellent problem in which four pairs of pieces: Re2/Bc3; Sg3/Sb5; Pe6/Pf6; Pc6/Pd6 reciprocally change their functions. The Zilahi theme here is presented with great lightness and economy. The finals with chameleon-echo mates are beautiful and instructive! Such a complex problem undoubtedly deserves admiration and a prize! But why "only" a special prize? My arguments are this: despite the extremely rich content, the problem has some minor minuses: there is some symmetry in the play, and the limited material suggests that both the Kings (far from each other) should practically be transformed into a Pawn phase. But again I repeat my overall assessment - this is an excellent problem!
FEN: 8/2K5/2ppPP2/1N6/8/2b3N1/3kr3/8
(C+ by Popeye)
1st Hon. Mention
Borislav GADJANSKI, SRB
white Qg6 Pd7a6 Kg5 Rb7 Sd3
black Pd5 Kc6 Rd6 Sh4
hs#3
2 sol.
(6+4)
Frankfurt Chess
1.Qg6-e4 Rd6*d7=P 2.Kg5*h4=rS d7-d6 3.rSh4-g2 d5*e4=Q#
1.Qg6-e6 Sh4-f5 2.Kg5*f5=rS d5-d4 3.rSf5-h6 Rd6*e6=Q #
A nice and non-standard problem in which the play is dictated only by the Frankfurt Chess condition. The most interesting point here is that the third moves of the white (W3) are without checks but provoking black pins and zugzwang. The model mates are another plus here although the play does not have a complete analogy. (An Important note: If we use the Protean Chess condition in the diagram position, the problem will not have a solution after 1.Qe4 Rxd7=P because here the move 2 ... Pd7-d6 would be illegal!).
FEN: 8/1R1P4/P1kr2Q1/3p2K1/7n/3N4/8/8
(C+ by WinChloe)
2nd Hon. Mention
Vlaicu CRISAN & S.K. BALASUBRAMANIAN, ROU/IND
white Bg1 Re7
white Royal Sd4
black Qh5 Rf2 Ka3 Bd7
h#3
2 sol.
(3+4)
Protean Chess
Royal Sd4
1.Bg1*f2=R Qh5-e8 2.Re7*e8=Q Bd7-g4 3.Qe8-a4 + Ka3*a4=rQ #
1.Re7*d7=B Qh5-h2 2.Bg1*h2=Q Rf2-f6 3.Qh2-b2 + Ka3*b2=rQ #
A very pleasant Miniature/Aristocrat in which it is not easy to find solutions. The moves of both sides elegantly demonstrate the reciprocal change of functions between them. A further plus are the model mates and ODT, but sadly the the Royal Sd4 remains stationary until the end.
FEN: 8/3bR3/8/7q/3N4/k7/5r2/6B1
(C+ by Popeye + WinChloe)
3rd Hon. Mention
Luis Miguel MARTÍN, ESP
white Pb7h6 Rg6f4 Ba8c3 Kg5
black Re2h1 Ka7 Bd7b4 Sh5
hs#3
2 sol.
(7+6)
Protean Chess
1.Bc3-e5 Re2*e5=B 2.Rf4-f5 Be5-b8 3.Rf5-a5 + Bb4*a5=R #
1.Rf4-f8 Bb4*f8=R 2.Bc3-d2 Rf8-b8 3.Bd2-e3 + Re2*e3=B #
Here it is obvious that white has to control the b8 square and the achieving this goal seems easy at first glance: such an idea is possible to realize using the wRf4 or wBc3 as control pieces. In reality, however, this idea turns out to be wrong! The b8 square must be blocked by the bRe2 or bBc3 after the sacrifices of wRf4/wBc3 (Annihilation!). Although the blocks on b8 have an orthodox character, the paradoxical content of this problem deserves a good evaluation!
FEN: B7/kP1b4/6RP/6Kn/1b3R2/2B5/4r3/7r
(C+ by Popeye + WinChloe)
Commendation
Juraj LÖRINC, SVK
white Kb2 Rc8b8
black Pg6h6c4b4 Kh7 Ba2 Sc2
hs#3
2 sol.
(3+7)
Frankfurt Chess
1.Kb2*a2=rB Sc2-d4 2.rBa2*c4=rP Sd4-b5 3.Rc8-c7 + Sb5*c7=R #
1.Kb2*c2=rS Ba2-b1 + 2.rSc2*b4=rP Bb1-e4 3.Rb8-b7 + Be4*b7=R #
This play of three pairs of thematic pieces (with reciprocal change of functions), ending with two chameleon-echo mates deserves attention. But, as I have already noted in my commentary for the problem awarded with a special prize, in analogous mechanisms the transformation of the wK into Pawn phases is obvious, and there also exists some degree of sketchiness and symmetry in the solutions.
FEN: 1RR5/7k/6pp/8/1pp5/8/bKn5/8
(C+ by WinChloe)
Commendation
Karol MLYNKA, SVK
white Rg3 Bd1
white Royal Pg2
black Royal Ph2
hs#3
2 sol.
(3+1)
Protean Chess
Royal Pg2, Ph2
1.Bd1-c2 rPh2-h1=rR 2.Bc2-h7 rRh1*h7=rB 3.Rg3-g6 rBh7*g6=rR #
1.Rg3-g5 rPh2-h1=rB + 2.rPg2-g4 rBh1-a8 3.Bd1-f3 + rBa8*f3 #
A fun play using two Royal pawns. For a problem with just 4 pieces (it is so called Wenigsteiner), the content looks good, although it does not have a thematic identity.
FEN: 8/8/8/8/8/6R1/6Pp/3B4
(C+ by Popeye + WinChloe)
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